In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

First, you would find the change in energy (ΔE). The equations we know are: c = λv and E = hv. Since the wavelength is given, you could find E via E = hc / λ.

Therefore: E = (6.626 x 10

^{-34} J s)(2.998 x 10

^{8} m/s) / (102.6 x 10

^{-9} m) = 1.936 x 10

^{-18} J

It's important to realize that energy is being

emitted, which means the e- is transitioning from a higher to lower energy level, so ΔE < 0. (The energy of electrons can be either positive or negative, but the energy of photons emitted/absorbed must always be positive.) Therefore, ΔE = -1.936 x 10-

^{18} J.

The problem indicates that the line is observed in the UV spectrum, so it deals with the Lyman series, which ends at n = 1 (n

_{final} = 1).

E

_{final} = -hR / n

_{final}^{2} = -(6.626 x 10

^{-34} J s)(3.28984 x 10

^{15} Hz) / 1

^{2} = -2.180 x 10

^{-18} J

We can then plug in the values we found into the equation ΔE = E

_{final} - E

_{initial}. So:

-1.936 x 10

^{-18} J = -2.180 x 10

^{-18} J - E

_{initial}E

_{initial} = -2.180 x 10

^{-18} J - (-1.936 x 10

^{-18} J)

E

_{initial} = -2.438 x 10

^{-19} J = -hR / n

_{initial}^{2}n

_{initial}^{2} = -hR / -2.438 x 10

^{-19} J = - (6.626 x 10

^{-34} J s)(3.28984 x 10

^{15} Hz) / (-2.438 x 10

^{-19} J)

n

_{initial} = 3, so the e- transitions down from the energy level n = 3 to n = 1.